A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s^2. The free-body diagram shows the forces acting on the cart. A free body diagram with 3 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline. The second vector is pointing up, labeled F Subscript N Baseline. The third vector is pointing up to the right at an angle of 35 degrees, labeled F Subscript p Baseline = 250 N. The up and down vectors are the same length.

The mass of the cart, to the nearest whole number, is

Split up the acceleration into horizontal and vertical components:

a = (1.4 m/s²) (cos(35º) i + sin(35º) j)

a ≈ (1.15 m/s²) i + (0.803 m/s²) j

Do the same with the pulling force (denoted by p):

p = (250 N) (cos(35º) i + sin(35º) j)

p ≈ (205 N) i + (143 N) j

By Newton's second law, the net horizontal force is proportional to the horizontal acceleration according to

∑ F = m a

The only horizontal force acting on the cart is the horizontal component of the pulling force. So we have

205 N = m (1.15 m/s²)

m ≈ 178.571 kg ≈ 180 kg

c.the charges if the proton and electron and gravity

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2